How do you find the $lim_(xrarr3) (x^2+x-12)/(x-3)$ ?

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Answer 1 · 2017-04-16T01:34:48

$7$

Rewrite the numerator as $(x+4)(x-3)$

$lim_(x->3) ((x+4)(x-3))/(x-3)$

Cancel the $x-3$

$lim_(x->3) x+4$

Plug in $3$ for $x$ :

$3+4=7$

Answer 2 · 2017-04-16T01:47:38

The numerator factors and cancels the denominator, leaving a simple linear expression that can be evaluated at the limit.

Given: $lim_(xrarr3) (x^2+x-12)/(x-3)$

Factor the numerator:

$lim_(xrarr3) ((x-3)(x+4))/(x-3)$

Please observe the factors that cancel:

$lim_(xrarr3) (cancel(x-3)(x+4))/cancel(x-3)$

This leaves a simple linear expression that can be evaluated at the limit:

$lim_(xrarr3) x+4 = 7$

Answer 3 · 2017-04-16T18:47:55

$7$

$"factorise numerator and simplify"$

$lim_(xto3)((x+4)(cancel(x-3)))/(cancel(x-3))$

$=lim_(xto3)(x+4)=3+4=7$

How do you find the lim_(xrarr3) (x^2+x-12)/(x-3)lim_(xrarr3) (x^2+x-12)/(x-3)?